Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 25, Problem 115P

(a)

To determine

The time constant for the charging of capacitor.

(a)

Expert Solution
Check Mark

Answer to Problem 115P

The time constant for the charging of capacitor is 10.0ms .

Explanation of Solution

Given:

The given values in the circuit are,

  ε=800VVc=4.20VR2=1.00mΩR1=0.500MΩ

And,

  C=20.0nF

Formula Used:

The expression for the time constant for charging the capacitor is given by,

  τ=R1C

Calculation:

The time constant for charging the capacitor is calculated as,

  τ=R1C=(0.500MΩ× 10 6 Ω 1MΩ)(20.0nF× 10 9 F 1nF)=10.0×103s× 103ms1s=10.0ms

Conclusion:

Therefore, the time constant for the charging of capacitor is 10.0ms .

(b)

To determine

The proof that the potential across switch increases from 0.200V to 4.20V .

(b)

Expert Solution
Check Mark

Answer to Problem 115P

The proof that as the potential across switch increases from 0.200V to 4.20V , the potential across the capacitor increases linearly with time is stated below.

Explanation of Solution

Given:

The given values in the circuit are,

  ε=800VVc=4.20VR2=1.00mΩR1=0.500MΩ

And,

  C=20.0nF

Formula Used:

The expression for the voltage across the charging capacitor is given by,

  V(t)=ε(1e t/τ )et/τ=1V(t)ε ....... (I)

Now, V(t)<<ε .

Let ,

  η=V(t)ε

Rewrite equation (I) as,

  et/τ=1ηet/τ=(1η)11+η ....... (II)

Expand the term et/τ as,

  et/τ=1+1τ+12!( t τ)2+...1+1τt

Equation (II) can be written as,

  1+1τt=1+η=1+V(t)εV(t)=ετt

This is a linear function.

Conclusion:

Therefore, the proof that as the potential across switch increases from 0.200V to 4.20V , the potential across the capacitor increases linearly with time is stated above.

(c)

To determine

The changed value of R1 so that the capacitor charges from 0.200V to 4.20V in 0.100s .

(c)

Expert Solution
Check Mark

Answer to Problem 115P

The changed value of R1 so that the capacitor charges from 0.200V to 4.20V in 0.100s is 1.00GΩ .

Explanation of Solution

Given:

The given values in the circuit are,

  ε=800VVc=4.20VR2=1.00mΩR1=0.500MΩ

And,

  C=20.0nF

Formula Used:

The expression for the potential across the capacitor is given by

  ΔV(t)=ετΔt=εR1CΔtR1=εCΔV(t)Δt

Calculation:

The changed value of R1 is calculated as,

  R1=εCΔV(t)Δt=( 800V)( 0.10s)( 20.0nF)( 4.2V0.2V)=1×109Ω× 10 91Ω=1.00GΩ

Conclusion:

Therefore, the changed value of R1 so that the capacitor charges from 0.200V to 4.20V in 0.100s is 1.00GΩ .

(d)

To determine

The time elapse during the discharge of capacitor when the switch S closes.

(d)

Expert Solution
Check Mark

Answer to Problem 115P

The time elapse during the discharge of capacitor when the switch S closes is 60.9ps .

Explanation of Solution

Given:

The given values in the circuit are,

  ε=800VVc=4.20VR2=1.00mΩR1=0.500MΩ

And,

  C=20.0nF

Formula Used:

The expression for the potential difference across the capacitor is given by,

  VC(t)=Vc0et/τ ....... (III)

The expression for τ is given by,

  τ=R2C

Rewrite equation (III) as,

  VC(t)=Vc0et/τVC(t)=Vc0et/ R 2Ct=R2Cln( V C ( t ) V C0 )

Calculation:

The time elapsed during the discharge of capacitor is calculated as,

  t=R2Cln( V C ( t ) V C0 )=(0.001Ω)(20.0nF)ln( 0.2V 4.2V)=0.06089×109s× 10 12ps1s=60.9ps

Conclusion:

Therefore, the time elapse during the discharge of capacitor when the switch S closes is 60.9ps .

(e)

To determine

The average rate at which the energy is delivered to resistor R1 and to discharge the resistance R2 .

(e)

Expert Solution
Check Mark

Answer to Problem 115P

The average rate at which the energy is delivered to resistor R1 is 6.17×109W and to switch the resistance R2 to discharge is 2.89kW .

Explanation of Solution

Given:

The given values in the circuit are,

  ε=800VVc=4.20VR2=1.00mΩR1=0.500MΩ

And,

  C=20.0nF

Formula Used:

The expression for the rate at which energy is dissipated in R1 is given by.

  P1=ΔE1Δt=I2R1

The current varies with time. So, integrate over time.

  ΔE1= I 2 R 1dt= ( V( t ) R 1 ) 2dt= ( εt τ R 1 ) 2 R 1dt=( ε τ)2(1 R 1 ) 0.005 0.105 t 2dt

The expression for the rate at which energy is dissipated in the switch resistance is given by,

  P2=12C(Vi2Vf2Δt)

Calculation:

The value of ΔE1 is calculated as,

  ΔE1=(ετ)(1 R 1 ) 0.005 0.105 t 2dt=( 800V 20s)2(1 1GΩ× 10 9 Ω 1GΩ )( t 3 3)0.0050.105=6.17×1010J

The value of P1 is calculated as,

  P1=ΔE1Δt=6.17× 10 10J0.1s=6.17×109W

The rate at which energy is dissipated in the switch resistance is calculated as,

  P2=12(20.0nF× 10 9 F 1nF)( ( 4.2V ) 2 ( 0.2V ) 2 60.9ps× 10 12 s 1s )=2.89×103W

Conclusion:

Therefore, the average rate at which the energy is delivered to resistor R1 is 6.17×109W and to switch the resistance R2 to discharge is 2.89kW .

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Chapter 25 Solutions

Physics for Scientists and Engineers

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